
A bird flying above a pond starts moving vertically down towards the water below at a speed of $ {\text{v m}}{{\text{s}}^{ - 1}} $ . Find its apparent velocity as viewed by a fish located at a depth of $ {\text{d m}} $ below the surface of water of refractive index $ \mu $ .
(A) $ \mu v $
(B) $ \dfrac{\mu }{v} $
(C) $ {\mu ^2}v $
(D) $ \mu {v^2} $
Answer
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Hint: To solve this question, we need to use the formula for the apparent depth of an object with respect to a transparent medium. Differentiating that expression with respect to the time, we will get the expression for the apparent velocity of the bird as viewed by the fish.
Formula used : The formula used to solve this question is given by
$ \dfrac{{h'}}{h} = \mu $ , here $ h' $ is the apparent height of an object as viewed by an observer inside a medium of refractive index $ \mu $ , and the real height of the object is equal to $ h $ .
Complete step by step answer:
Let the bird be at a vertical height of $ y $ above the surface of water at a time $ t $ .
So for the fish inside the pond, the height of the bird is given by
$ h = \mu y $ ………………………..(1)
Differentiating both sides of (1) with respect to the time $ t $ we get
$ \dfrac{{dh}}{{dt}} = \mu \dfrac{{dy}}{{dt}} $ ………………………..(2)
According to the question, the bird is moving vertically downwards towards the water at the speed of $ {\text{v m}}{{\text{s}}^{ - 1}} $ . So the rate of change of the height of the bird above the water surface is equal to $ {\text{v m}}{{\text{s}}^{ - 1}} $ , that is,
$ \dfrac{{dy}}{{dt}} = v $
Substituting this in (2) we get
$ \dfrac{{dh}}{{dt}} = \mu v $
The height $ h $ is equal to the apparent height of the bird as observed by the fish under water. So the apparent velocity of the bird is equal to the rate of change of the apparent height with respect to the time, $ \dfrac{{dh}}{{dt}} $ . So the apparent velocity of the bird is given as
$ v' = \mu v $
Thus, the apparent velocity of the bird as viewed by the fish under water is equal to $ \mu v $ .
Hence, the correct answer is option A.
Note<
We must note that the value of the depth of the fish given in this question is not needed in the solution of this problem. It is just the extra information given in this question. We must not be confused by this and must focus on the right approach for solving this question.
Formula used : The formula used to solve this question is given by
$ \dfrac{{h'}}{h} = \mu $ , here $ h' $ is the apparent height of an object as viewed by an observer inside a medium of refractive index $ \mu $ , and the real height of the object is equal to $ h $ .
Complete step by step answer:
Let the bird be at a vertical height of $ y $ above the surface of water at a time $ t $ .
So for the fish inside the pond, the height of the bird is given by
$ h = \mu y $ ………………………..(1)
Differentiating both sides of (1) with respect to the time $ t $ we get
$ \dfrac{{dh}}{{dt}} = \mu \dfrac{{dy}}{{dt}} $ ………………………..(2)
According to the question, the bird is moving vertically downwards towards the water at the speed of $ {\text{v m}}{{\text{s}}^{ - 1}} $ . So the rate of change of the height of the bird above the water surface is equal to $ {\text{v m}}{{\text{s}}^{ - 1}} $ , that is,
$ \dfrac{{dy}}{{dt}} = v $
Substituting this in (2) we get
$ \dfrac{{dh}}{{dt}} = \mu v $
The height $ h $ is equal to the apparent height of the bird as observed by the fish under water. So the apparent velocity of the bird is equal to the rate of change of the apparent height with respect to the time, $ \dfrac{{dh}}{{dt}} $ . So the apparent velocity of the bird is given as
$ v' = \mu v $
Thus, the apparent velocity of the bird as viewed by the fish under water is equal to $ \mu v $ .
Hence, the correct answer is option A.
Note<
We must note that the value of the depth of the fish given in this question is not needed in the solution of this problem. It is just the extra information given in this question. We must not be confused by this and must focus on the right approach for solving this question.
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